NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers

Exponents and Powers Class 8 Questions And Answers PDF Free Download

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Exponents and Powers Class 8 NCERT Solutions (Exercise)

Question:1 (i) Evaluate.

$3^{-2}$

Answer:

The detailed explanation for the above-written question is as follows,

We know that,

$a^{-m}=\frac{1}{a^{m}}$

So, here m =2 and a = 3

$3^{-2}=\frac{1}{3^{2}} = \frac{1}{3}\times \frac{1}{3} = \frac{1}{9}$

Question: 1(ii) Evaluate.

$(-4)^{-2}$

Answer:

The detailed explanation for the above-written question is as follows

We know that,

$a^{-m}= \frac{1}{a^{m}}$

So, here (a = -4) and (m = 2)

Then, according to the law of exponent

$(-4)^{-2}= \frac{1}{(-4)^{2}} = \frac{1}{(-4)}\times \frac{1}{(-4)} = \frac{1}{16}$ [ negative $\times$ negative = positive]

Question: 1(iii) Evaluate.

$\left(\frac{1}{2}\right )^{-5}$

Answer:

The detailed solution for the above-written question is as follows

We know that,

$(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}$ $\&$ $a^{-m} = \frac{1}{a^{m}}$

So, here

a = 1 and b = 2 and m =-5

According to the law of exponent

$(\frac{1}{2})^{-5}=\frac{1^{-5}}{2^{-5}} = \frac{-1}{2^{-5}}$

$=(-1)\times -2\times -2\times -2\times -2\times -2 = 32$

Question: 2(i) Simplify and express the result in power notation with a positive exponent.

$(-4)^{5}\div (-4)^{8}$

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

$\frac{a^{m}}{a_{n}} = a^{m-n}$ and $a^{-m}= \frac{1}{a^{m}}$

So according to this

a = -4, m = 5 and n = 8

$\frac{4^{5}}{4_{8}} = -4^{5-8} = -4^{-3}$

$= (-\frac{1}{4})\times -(\frac{1}{4})\times(-\frac{1}{4}) = -\frac{1}{64}$

Question: 2(ii) Simplify and express the result in power notation with a positive exponent.

$\left (\frac{1}{2^3} \right )^2$

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

$\frac{a^{m}}{b^{m}} = (\frac{a}{b})^{m}$ and $a^{-m}= \frac{1}{a^{m}}$ and $(a^{m})^{n} = a^{mn}$

So, we have given

a = 1, b=2

By using the above exponential law,

$\frac{a^{m}}{b^{m}} = \frac{1}{(2^{3})^{2}} = \frac{1}{2^{6}}$

$= \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} = \frac{1}{64}$

Question: 2(iii) Simplify and express the result in power notation with a positive exponent.

$(-3)^4\times \left(\frac{5}{3} \right )^{4}$

Answer:

The detailed solution for the above-written question is as follows,

We know the exponential formula

$(\frac{a}{b})^{m}= \frac{a^{m}}{b^{m}}$

So, $(-3)^{4}\times (\frac{5}{3})^{4}= (-3)^{4}\times \frac{5^{4}}{3^{4}} = \frac{625}{1}$

Question: 2(iv) Simplify and express the result in power notation with a positive exponent.

$(3^{-7}\div 3^{-10})\times 3^{-5}$

Answer:

The detailed explanation for the above-written question is as follows

As we know, the exponential form

$\frac{a^{m}}{b^{n}}= a^{m-n} \& (a^{m}\times a^{n})=a^{m+n}$

By using these two forms, we get,

$\frac{3^{-7}}{3^{-10}}\times (3)^{-5}=3^{-7-(-10)} \times (3)^{-5}$

$=3^{3} \times (3)^{-5}= 3^{-2}$

$=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

Question:2(v) Simplify and express the result in power notation with a positive exponent.

$2^{-3} \times (-7)^{-3}$

Answer:

The detailed solution for the above-written question is as follows,

we know the exponential forms

$a^{-m}=\frac{1}{a^{m}}$ & $a^{m}\times b^{m}= (ab)^{m}$

So, according to our data,

Here, initially, we use the first form and then the second one.

$2^{-3}\times (-7)^{-3}= \frac{1}{2^{3}}\times \frac{1}{(-7)^{3}}$

$= \frac{1}{(2\times -7)^{3}}= \frac{1}{(-14)^{3}}$

Question:3(i) Find the value of.

$(3^0 + 4^{-1})\times 2^2$

Answer:

The detailed explanation for the above-written question is as follows,

As we know that $a^{0}=1$

So, $3^{0}=1$

now,

$=(1+\frac{1}{4})\times 2^{2}$

$=\frac{5}{4}\times 2^{2}\Rightarrow \frac{5}{2^{2}}\times 2^{2}=5/1$

Question: 3(ii) Find the value of.

$(2^{-1}\times 4^{-1})\div 2^{-2}$

Answer:

The detailed explanation for the above-written question is as follows

Rewrite the equation

$(2^{-1}\times 4^{-1})\div 2^{-2}$ $= (2^{-1}\times 2^{-2})\div 2^{-2}$

$=(2^{-1+(-2)})\div 2^{-2}$ ................................. $a^{m}\times a^{n}= a^{(m+n)}$

$=(2^{-3})\div 2^{-2} = 2^{-3-(-2)}$ ........................ $a^{m}\div a^{n}= a^{(m-n)}$

$= 2^{-1}= \frac{1}{2}$

Question: 3(iii) Find the value of.

$\left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}$

Answer:

The detailed explanation for the above-written question is as follows,

This is the exponential form

$(a/b)^{m}= \frac{a^{m}}{b^{m}}$

So, $\left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}$

$=\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\frac{1}{4^{-2}}$ .......................using this form $a^{m}= \frac{1}{a^{m}}$

$=2^{2}+3^{2}+4^{2}$

= 4+9+16

= 29

Question: 3(iv) Find the value of.

$(3^{-1} + 4^{-1} + 5^{-1})^0$

Answer:

since we know that

$a^{0}=1$

$(3^{-1} + 4^{-1} + 5^{-1})^0$ $=1$

Question: 3(v) Find the value of.

$\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}$

Answer:

The detailed explanation for the above-written question is as follows

$\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}$

$=(-2/3)^{-2\times 2}$ .............. By using these form of exponential $(a^{m})^{n}=a^{mn}$

$(-2/3)^{-4}=(-3/2)^{4}$ ......... use this $a^{-m}=\frac{1}{a^{m}}$

$=\frac{81}{16}$

Question:4(i) Evaluate

$\frac{8^{-1}\times 5^{3}}{2^{-4}}$

Answer:

The detailed explanation for the above-written question is as follows

$\frac{8^{-1}\times 5^{3}}{2^{-4}}$

After rewriting the above equation, we get,

$\frac{2^{-3}\times 5^{3}}{2^{-4}}$ $=2^{-3-(-4)}\times 5^{3}$ ...........as we know that $\frac{a^{m}}{a^{n}}= a^{m-n}$

$\\=2^{1}\times 5^{3}\\ = 2\times 125 = 250$

An alternate method,

$= \frac{5^{3}}{2^{-4}\times 2^{3}}$

here you can use first $a^{-m}= \frac{1}{a^{m}}$ and after that use $a^{m}\times a^{n}= a^{m+n}$

Question: 4(ii) Evaluate

$(5^{-1}\times 2^{-1})\times 6^{-1}$

Answer:

The detailed explanation for the above-written question is as follows

We clearly see that this is in the form of $a^{-m}=\frac{1}{a^{m}}$

So, $(5^{-1}\times 2^{-1})\times 6^{-1}= (\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}$

$= \frac{1}{60}$

Question: 5 Find the value of $m$ for which

$5^m \div 5^{-3} = 5^5$

Answer:

We have,

$a^{m}\div a^{n}= a^{m-n}$

Here a = 5 and n =-3 and m-n = 5

therefore,

$5^{m}\div 5^{-3}= 5^{m+3} = 5^{5}$

By comparing both sides, we get,

m+3 = 5

m= 2

Question: 6(i) Evaluate

$\left \{\left (\frac{1}{3} \right )^{-1} - \left( \frac{1}{4} \right )^{-1} \right \}^{-1}$

Answer:

The detailed solution for the above-written question is as follows

$\left \{\left (\frac{1}{3} \right )^{-1} - \left( \frac{1}{4} \right )^{-1} \right \}^{-1}$

$= [(1\times 3)-(1\times 4)]^{-1}$ .............by using $a^{-m}= \frac{1}{a^{m}}$

$= [3-4]^{-1}$

$\\= [-1]^{-1}\\=-1$

Question: 6(ii) Evaluate

$\left ( \frac{5}{8} \right )^{-7}\times \left (\frac{8}{5}\right)^{-4}$

Answer:

The detailed solution for the above-written question is as follows

$\left ( \frac{5}{8} \right )^{-7}\times \left (\frac{8}{5}\right)^{-4}$

$=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}}$ ................ using the form $\frac{a^{m}}{b^{m}}= (a/b)^{m}$

$=5^{-7+4}\times 8^{-4+7}$ ............using $a^{m} \div a^{n} = a^{m-n}$

$= 5^{-3}\times 8^{+3}$

$\\= \frac{8^{3}}{5^{3}}\\\\=\frac{512}{125}$

Question: 7(i) Simplify

$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\;\;(t\neq 0)$

Answer:

The detailed solution for the above-written question is as follows

$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\;\;(t\neq 0)$

we can write $25= 5^{2}$

So, after rewriting the equation,

$\frac{5^{2}\times t^{-4}}{5^{-3}\times 10\times t^{-8}}$

$= \frac{5^{2+3}\times t^{-4+8}}{10}$ .................using the form $[a^{m}\div a^{n}= a^{m-n}]$

$= \frac{5^{5}\times t^{4}}{10}$ .............(By expanding, we get,)

$= \frac{625t^{4}}{2}$

Question: 7(ii) Simplify.

$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}$

Answer:

The detailed solution for the above-written question is

$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}$

we can write 125 = $5^{3}$ and $6^{-5}$ can be written as $(2\times 3)^{-5}$

Now, rewriting the equation, we get

$=\frac{3^{-5}\times 10^{-5}\times 5^{3}}{5^{-7}\times(2\times 3) ^{-5}}$

$=\frac{3^{-5}\times 10^{-5}\times 5^{3+7}}{(2\times 3) ^{-5}}$ .............by using $[a^{m}\div a^{n}=a^{m-n}]$

$=\frac{ 10^{-5}\times 5^{10}}{(2) ^{-5}}$ .....................Use $[a^{m}\div a^{n}=a^{m-n}]$

$5^{10-5}= 5^{5}=3125$

As $[10^{-5} = (2\times 5)^{-5}=2^{-5}\times 5^{-5}]$ .

$2^{-5}$ can be cancelled out with the denominator $2^{-5}$

Question: 1(i) Express the following numbers in standard form.

0.0000000000085

Answer:

The standard form is $8.5\times 10^{-12}$

Question: 1(ii) Express the following numbers in standard form.

0.00000000000942

Answer:

The standard form is $9.42\times 10^{-12}$

Question: 1(iii) Express the following numbers in standard form.

6020000000000000

Answer:

The standard form is $6.02\times 10^{15}$

Question: 1(iv) Express the following numbers in standard form.

0.00000000837

Answer:

The standard form of the given number is $8.37\times10^{-9}$

Question: 1(v) Express the following numbers in standard form.

31860000000

Answer:

The standard form is $3.186\times10^{10}$

Question: 2(i) Express the following numbers in the usual form.

$3.02\times 10^{-6}$

Answer:

$3.02\times 10^{-6}$

$=\frac{3.02}{1000000}$

$=0.00000302$

This is the usual form.

Question: 2 (ii) Express the following numbers in the usual form.

$4.5 \times 10^4$

Answer:

$4.5 \times 10^4=4.5\times 10000$

$=45000$

This is the usual form.

Question: 2(iii) Express the following numbers in the usual form.

$3\times 10^{-8}$

Answer:

$3\times 10^{-8}$

$\frac{3}{100000000} = 0.000000030$

This is the usual form.

Question: 2(iv) Express the following numbers in the usual form.

$1.0001\times 10^9$

Answer:

$1.0001\times 10^9$

$=1.0001\times 100000000$

$=1000100000$

This is the usual form.

Question: 2(v) Express the following numbers in the usual form.

$5.8\times 10^{12}$

Answer:

$5.8\times 10^{12}$

$=5.8\times 100000000000$

$=5800000000000$

This is the usual form.

Question: 2(vi) Express the following numbers in the usual form.

$3.61492 \times 10^6$

Answer:

$3.61492 \times 10^6$

$= 3.61492\times 1000000$

$= 3614920$

17155

Question: 3(i) Express the number appearing in the following statements in standard form.

1 micron is equal to $\frac{1}{1000000}m$.

Answer:

1 micron is equal to

$\frac{1}{1000000}m$

$= 1\times 10^{-6}$

Question: 3(ii) Express the number appearing in the following statements in standard form.

The charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

Answer:

The charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

$=1.6\times 10^{-19}$ coulomb.

Question: 3(iii) Express the number appearing in the following statements in standard form.

The size of a bacteria is 0.0000005 m.

Answer:

The size of a bacteria is 0.0000005 m

$\frac{5}{10000000}=5\times 10^{-7}m$

Question: 3(iv) Express the number appearing in the following statements in standard form.

The size of a plant cell is 0.00001275 m.

Answer:

The size of a plant cell is 0.00001275 m

$=\frac{1275}{10000}=1.275\times10^{-5}m$

Question: 3(v) Express the number appearing in the following statements in standard form.

The thickness of a thick paper is 0.07 mm.

Answer:

The thickness of a thick paper is 0.07

$=\frac{7}{100}=7\times 10^{-2}mm$

Question: 4 In a stack, there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Answer:

The thickness of each book = 20mm

So, the thickness of 5 books = $(5\times 20)=100 mm$

The thickness of one paper sheet =0.016mm

So, the thickness of 5 paper sheet = $(5\times 0.016)=0.08 mm$

The total thickness of the stack = (100+0.08)mm

=100.08 mm or $(1.008\times10^{-2}mm)$

NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers - Topics

The topics discussed in the NCERT Solutions for class 8, chapter 10, Exponents and Powers, are:

• Powers with Negative Exponents
• Laws of Exponents
• Use of Exponents to Express Small Numbers in Standard Form
• Comparing very large and very small numbers

Exponents and Powers Class 8 Solutions - Important Formulae

Here are some important formulas or laws that will help students to solve problems related to Exponents and Powers.

• Law of Product: $a^m \times a^n=a^{(m+n)}$

• Law of Quotient: $\frac{a^m}{a^n}=a^{(m-n)}$

• Law of Zero Exponent: $a^{0} = 1$

• Law of Negative Exponent: $a^{(-m)}= \frac{1}{ a^m}$

• Law of Power of a Power: $\left(a^m\right)^n=a^{\left(mn\right)}$

• Law of Power of a Product: $(a b)^n=a^{n} b^n$

• Law of Power of a Quotient: $(\frac{a}{b})^m= \frac{a^m}{b^m}$

NCERT Solutions for Class 8 Maths - Chapter Wise

Given below are all the chapter-wise NCERT class 8 maths solutions given in one place for ease of access:

NCERT Solutions for Class 8 - Subject Wise

The NCERT solutions for class 8 Maths is one of the essential study materials for students in exam preparation. Check out the Subject-wise NCERT Solutions for Class 8 below.

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

NCERT Books and NCERT Syllabus

It is very important to focus on other study resources like NCERT Books and NCERT Syllabus to get an overview of the topics covered in this academic year. Students can use the below links to access the NCERT Syllabus and book PDFs.

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8